Tag: permutation representation

permutation representations of monodromy groups

Published March 20, 2007 by lieven

Today we will explain how curves defined over
$\overline{\mathbb{Q}} $ determine permutation representations
of the carthographic groups. We have seen that any smooth projective
curve $C $ (a Riemann surface) defined over the algebraic
closure $\overline{\mathbb{Q}} $ of the rationals, defines a
_Belyi map_ $\xymatrix{C \ar[rr]^{\pi} & & \mathbb{P}^1} $ which is only ramified over the three points
$\\{ 0,1,\infty \\} $. By this we mean that there are
exactly $d $ points of $C $ lying over any other point
of $\mathbb{P}^1 $ (we call $d $ the degree of
$\pi $) and that the number of points over $~0,1~ $ and
$~\infty $ is smaller than $~d $. To such a map we
associate a _dessin d\’enfant_, a drawing on $C $ linking the
pre-images of $~0 $ and $~1 $ with exactly $d $
edges (the preimages of the open unit-interval). Next, we look at
the preimages of $~0 $ and associate a permutation
$\tau_0 $ of $~d $ letters to it by cycling
counter-clockwise around these preimages and recording the edges we
meet. We repeat this procedure for the preimages of $~1 $ and
get another permutation $~\tau_1 $. That is, we obtain a
subgroup of the symmetric group $ \langle \tau_0,\tau_1
\rangle \subset S_d $ which is called the monodromy
group of the covering $\pi $.

For example, the
dessin on the right is
associated to a degree $8 $ map $\mathbb{P}^1 \rightarrow
\mathbb{P}^1 $ and if we let the black (resp. starred) vertices be
the preimages of $~0 $ (respectively of $~1 $), then the
corresponding partitions are $\tau_0 = (2,3)(1,4,5,6) $
and $\tau_1 = (1,2,3)(5,7,8) $ and the monodromy group
is the alternating group $A_8 $ (use
GAP ).

But wait! The map is also
ramified in $\infty $ so why don\’t we record also a
permutation $\tau_{\infty} $ and are able to compute it from
the dessin? (Note that all three partitions are needed if we want to
reconstruct $C $ from the $~d $ sheets as they encode in
which order the sheets fit together around the preimages). Well,
the monodromy group of a $\mathbb{P}^1 $ covering ramified only
in three points is an epimorphic image of the fundamental
group of the sphere
minus three points $\pi_1(\mathbb{P}^1 – { 0,1,\infty
}) $ That is, the group of all loops beginning and
ending in a basepoint upto homotopy (that is, two such loops are the
same if they can be transformed into each other in a continuous way
while avoiding the three points).

This group is generated by loops
$\sigma_i $ running from the basepoint to nearby the i-th
point, doing a counter-clockwise walk around it and going back to be
basepoint $Q_0 $ and the epimorphism to the monodromy group is given by sending

$\sigma_1 \mapsto \tau_0~\quad~\sigma_2 \mapsto
\tau_1~\quad~\sigma_3 \mapsto \tau_{\infty} $

Now,
these three generators are not independent. In fact, this fundamental
group is

$\pi_1(\mathbb{P}^1 – \\{ 0,1,\infty \\}) =
\langle \sigma_1,\sigma_2,\sigma_3~\mid~\sigma_1 \sigma_2
\sigma_3 = 1 \rangle $

To understand this, let us begin
with an easier case, that of the sphere minus one point. The fundamental group of the plane minus one point is
$~\mathbb{Z} $ as it encodes how many times we walk around the
point. However, on the sphere the situation is different as we can make
our walk around the point longer and longer until the whole walk is done
at the backside of the sphere and then we can just contract our walk to
the basepoint. So, there is just one type of walk on a sphere minus one
point (upto homotopy) whence this fundamental group is trivial. Next,
let us consider the sphere minus two points

Repeat the foregoing to the walk $\sigma_2 $, that
is, strech the upper part of the circular tour all over the backside of
the sphere and then we see that we can move it to fit with the walk
$\sigma_1$ BUT for the orientation of the walk! That is, if we do this
modified walk $\sigma_1 \sigma_2^{\’} $ we just made the
trivial walk. So, this fundamental group is $\langle
\sigma_1,\sigma_2~\mid~\sigma_1 \sigma_2 = 1 \rangle =
\mathbb{Z} $ This is also the proof of the above claim. For,
we can modify the third walk $\sigma_3 $ continuously so that
it becomes the walk $\sigma_1 \sigma_2 $ but
with the reversed orientation ! As $\sigma_3 =
(\sigma_1 \sigma_2)^{-1} $ this allows us to compute the
\’missing\’ permutation $\tau_{\infty} = (\tau_0
\tau_1)^{-1} $ In the example above, we obtain
$\tau_{\infty}= (1,2,6,5,8,7,4)(3) $ so it has two cycles
corresponding to the fact that the dessin has two regions (remember we
should draw ths on the sphere) : the head and the outer-region. Hence,
the pre-images of $\infty$ correspond to the different regions of the
dessin on the curve $C $. For another example,
consider the degree 168 map

$K \rightarrow \mathbb{P}^1 $

which is the modified orbit map for the action of
$PSL_2(\mathbb{F}_7) $ on the Klein quartic.
The corresponding dessin is the heptagonal construction of the Klein
quartic

Here, the pre-images of 1 correspond to the midpoints of the
84 edges of the polytope whereas the pre-images of 0 correspond to the
56 vertices. We can label the 168 half-edges by numbers such that
$\tau_0 $ and $\tau_1 $ are the standard generators b
resp. a of the 168-dimensional regular representation (see the atlas
page ).
Calculating with GAP the element $\tau_{\infty} = (\tau_0
\tau_1)^{-1} = (ba)^{-1} $ one finds that this permutation
consists of 24 cycles of length 7, so again, the pre-images of
$\infty $ lie one in each of the 24 heptagonal regions of the
Klein quartic. Now, we are in a position to relate curves defined
over $\overline{Q} $ via their Belyi-maps and corresponding
dessins to Grothendiecks carthographic groups $\Gamma(2) $,
$\Gamma_0(2) $ and $SL_2(\mathbb{Z}) $. The
dessin gives a permutation representation of the monodromy group and
because the fundamental group of the sphere minus three
points $\pi_1(\mathbb{P}^1 – \\{ 0,1,\infty \\}) =
\langle \sigma_1,\sigma_2,\sigma_3~\mid~\sigma_1 \sigma_2
\sigma_3 = 1 \rangle = \langle \sigma_1,\sigma_2
\rangle $ is the free group op two generators, we see that
any dessin determines a permutation representation of the congruence
subgroup $\Gamma(2) $ (see this
post where we proved that this
group is free). A clean dessin is one for which one type of
vertex has all its valancies (the number of edges in the dessin meeting
the vertex) equal to one or two. (for example, the pre-images of 1 in
the Klein quartic-dessin or the pre-images of 1 in the monsieur Mathieu
example ) The corresponding
permutation $\tau_1 $ then consists of 2-cycles and hence the
monodromy group gives a permutation representation of the free
product $C_{\infty} \ast C_2 =
\Gamma_0(2) $ Finally, a clean dessin is said to be a
quilt dessin if also the other type of vertex has all its valancies
equal to one or three (as in the Klein quartic or Mathieu examples).
Then, the corresponding permutation has order 3 and for these
quilt-dessins the monodromy group gives a permutation representation of
the free product $C_2 \ast C_3 =
PSL_2(\mathbb{Z}) $ Next time we will see how this lead
Grothendieck to his anabelian geometric approach to the absolute Galois
group.

The cartographers’ groups

Published March 13, 2007 by lieven

Just as cartographers like
Mercator drew maps of
the then known world, we draw dessins
d ‘enfants to depict the
associated algebraic curve defined over
$\overline{\mathbb{Q}} $.

In order to see that such a dessin
d’enfant determines a permutation representation of one of
Grothendieck’s cartographic groups, $SL_2(\mathbb{Z}),
\Gamma_0(2) $ or $\Gamma(2) $ we need to have realizations of these
groups (as well as their close relatives
$PSL_2(\mathbb{Z}),GL_2(\mathbb{Z}) $ and $PGL_2(\mathbb{Z}) $) in
terms of generators and relations.

As this lesson will be rather
technical I’d better first explain what we will prove (so that you can
skip it if you feel comfortable with the statements) and why we want to
prove it. What we will prove in detail below is that these groups
can be written as free (or amalgamated) group products. We will explain
what this means and will establish that

$PSL_2(\mathbb{Z}) = C_2
\ast C_3, \Gamma_0(2) = C_2 \ast C_{\infty}, \Gamma(2)
= C_{\infty} \ast C_{\infty} $

$SL_2(\mathbb{Z}) =
C_4 \ast_{C_2} C_6, GL_2(\mathbb{Z}) = D_4 \ast_{D_2} D_6,
PGL_2(\mathbb{Z}) = D_2 \ast_{C_2} D_3 $

where $C_n $ resp.
$D_n $ are the cyclic (resp. dihedral) groups. The importance of these
facts it that they will allow us to view the set of (isomorphism classes
of) finite dimensional representations of these groups as
noncommutative manifolds . Looking at the statements above we
see that these arithmetical groups can be build up from the first
examples in any course on finite groups : cyclic and dihedral
groups.

Recall that the cyclic group of order n, $C_n $ is the group of
rotations of a regular n-gon (so is generated by a rotation r with
angle $\frac{2 \pi}{n} $ and has defining relation $r^n = 1 $, where 1
is the identity). However, regular n-gons have more symmetries :
flipping over one of its n lines of symmetry

The dihedral group $D_n $ is the group generated by the n
rotations and by these n flips. If, as before r is a generating
rotation and d is one of the flips, then it is easy to see that the
dihedral group is generated by r and d and satisfied the defining
relations

$r^n=1 $ and $d^2 = 1 = (rd)^2 $

Flipping twice
does nothing and to see the relation $~(rd)^2=1 $ check that doing twice a
rotation followed by a flip brings all vertices back to their original
location. The dihedral group $D_n $ has 2n elements, the n-rotations
$r^i $ and the n flips $dr^i $.

In fact, to get at the cartographic
groups we will only need the groups $D_4, D_6 $ and their
subgroups. Let us start by finding generators of the largest
group $GL_2(\mathbb{Z}) $ which is the group of all invertible $2
\times 2 $ matrices with integer coefficients.

Consider the
elements

$U = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix},
V = \begin{bmatrix} 0 & 1 \\ -1 & 1 \end{bmatrix}/tex] and $R =
\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} $

and form the
matrices

$X = UV = \begin{bmatrix} 1 & -1 \\ 0 & 1
\end{bmatrix}, Y = VU = \begin{bmatrix} 1 & 0 \\ 1 & 1
\end{bmatrix} $

By induction we prove the following relations in
$GL_2(\mathbb{Z}) $

$X^n \begin{bmatrix} a & b \\ c & d
\end{bmatrix} = \begin{bmatrix} a-nc & b-nd \\ c & d \end{bmatrix} $
and $\begin{bmatrix} a & b \\ c& d \end{bmatrix} X^n =
\begin{bmatrix} a & b-na \\ c & d-nc \end{bmatrix} $

$Y^n \begin{bmatrix} a & b \\ c & d \end{bmatrix} =
\begin{bmatrix} a & b \\ c+na & d+nb \end{bmatrix} $ and
$\begin{bmatrix} a & b \\ c & d \end{bmatrix} Y^n = \begin{bmatrix}
a+nb & b \\ c+nd & d \end{bmatrix} $

The determinant ad-bc of
a matrix in $GL_2(\mathbb{Z}) $ must be $\pm 1 $ whence all rows and
columns of

$\begin{bmatrix} a & b \\ c & d \end{bmatrix} \in
GL_2(\mathbb{Z}) $

consist of coprime numbers and hence a and
c can be reduced modulo each other by left multiplication by a power
of X or Y until one of them is zero and the other is $\pm 1 $. We
may even assume that $a = \pm 1 $ (if not, left multiply with U).

So,
by left multiplication by powers of X and Y and U we can bring any
element of $GL_2(\mathbb{Z}) $ into the form

$\begin{bmatrix}
\pm 1 & \beta \\ 0 & \pm 1 \end{bmatrix} $

and again by left
multiplication by a power of X we can bring it in one of the four
forms

$\begin{bmatrix} \pm 1 & 0 \\ 0 & \pm 1 \end{bmatrix}
= { 1,UR,RU,U^2 } $

This proves that $GL_2(\mathbb{Z}) $ is
generated by the elements U,V and R.

Similarly, the group
$SL_2(\mathbb{Z}) $ of all $2 \times 2 $ integer matrices with
determinant 1 is generated by the elements U and V as using the
above method and the restriction on the determinant we will end up with
one of the two matrices

${ \begin{bmatrix} 1 & 0 \\ 0 & 1
\end{bmatrix},\begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} } =
{ 1,U^2 } $

so we never need the matrix R. As for
relations, there are some obvious relations among the matrices U,V and
R, namely

$U^2=V^3 $ and $1=U^4=R^2=(RU)^2=(RV)^2$ $

The
real problem is to prove that all remaining relations are consequences
of these basic ones. As R clearly has order two and its commutation
relations with U and V are just $RU=U^{-1}R $ and $RV=V^{-1}R $ we can
pull R in any relation to the far right and (possibly after
multiplying on the right with R) are left to prove that the only
relations among U and V are consequences of $U^2=V^3 $ and
$U^4=1=V^6 $.

Because $U^2=V^3 $ this element is central in the
group generated by U and V (which we have seen to be
$SL_2(\mathbb{Z}) $) and if we quotient it out we get the modular
group

$\Gamma = PSL_2(\mathbb{Z}) $

Hence in order to prove our claim
it suffices that

$PSL_2(\mathbb{Z}) = \langle
\overline{U},\overline{V} : \overline{U}^2=\overline{V}^3=1
\rangle $

Phrased differently, we have to show that
$PSL_2(\mathbb{Z}) $ is the free group product of the cyclic groups of
order two and three (those generated by $u = \overline{U} $ and
$v=\overline{V} $) $C_2 \ast C_3 $

Any element of this free group
product is of the form $~(u)v^{a_1}uv^{a_2}u \ldots
uv^{a_k}(u) $ where beginning and trailing u are optional and
all $a_i $ are either 1 or 2.

So we have to show that in
$PSL_2(\mathbb{Z}) $ no such word can give the identity
element. Today, we will first sketch the classical argument based
on the theory of groups acting on trees due to Jean-Pierre
Serre and Hyman Bass. Tomorrow, we will give a short elegant proof due to
Roger Alperin and draw
consequences to the description of the carthographic groups as
amalgamated free products of cyclic and dihedral groups.

Recall
that $GL_2(\mathbb{Z}) $ acts via Moebius
transformations on
the complex plane $\mathbb{C} = \mathbb{R}^2 $ (actually it is an
action on the Riemann sphere $\mathbb{P}^1_{\mathbb{C}} $) given by the
maps

$\begin{bmatrix} a & b \\ c & d \end{bmatrix}.z =
\frac{az+b}{cz+d} $

Note that the action of the
center of $GL_2(\mathbb{Z}) $ (that is of $\pm \begin{bmatrix} 1 & 0
\\ 0 & 1 \end{bmatrix} $) acts trivially, so it is really an action of
$PGL_2(\mathbb{Z}) $.

As R interchanges the upper and lower half-plane
we might as well restrict to the action of $SL_2(\mathbb{Z}) $ on the
upper-halfplane $\mathcal{H} $. It is quite easy to see that a
fundamental domain
for this action is given by the greyed-out area

To see that any $z \in \mathcal{H} $ can be taken into this
region by an element of $PSL_2(\mathbb{Z}) $ note the following two
Moebius transformations

$\begin{bmatrix} 1 & 1 \\ 0 & 1
\end{bmatrix}.z = z+1 $ and $\begin{bmatrix} 0 & 1 \\ -1
& 0 \end{bmatrix}.z = -\frac{1}{z} $

The first
operation takes any z into a strip of length one, for example that
with Re(z) between $-\frac{1}{2} $ and $\frac{1}{2} $ and the second
interchanges points within and outside the unit-circle, so combining the
two we get any z into the greyed-out region. Actually, we could have
taken any of the regions in the above tiling as our fundamental domain
as they are all translates of the greyed-out region by an element of
$PSL_2(\mathbb{Z}) $.

Of course, points on the boundary of the
greyed-out fundamental region need to be identified (in order to get the
identification of $\overline{\mathcal{H}/PSL_2(\mathbb{Z})} $ with the
Riemann sphere $S^2=\mathbb{P}^1_{\mathbb{C}} $). For example, the two
halves of the boundary by the unit circle are interchanged by the action
of the map $z \rightarrow -\frac{1}{z} $ and if we take the translates under
$PSL_2(\mathbb{Z}) $ of the indicated circle-part

we get a connected tree with fundamental domain the circle
part bounded by i and $\rho = \frac{1}{2}+\frac{\sqrt{3}}{2} i $.
Calculating the stabilizer subgroup of i (that is, the subgroup of
elements fixing i) we get that this subgroup
is $\langle u \rangle = C_2 $ whereas the stabilizer subgroup of
$\rho $ is $\langle v \rangle = C_3 $.

Using this facts and the general
results of Jean-Pierre Serres book Trees
one deduces that $PSL_2(\mathbb{Z}) = C_2 \ast C_3 $
and hence that the obvious relations among U,V and R given above do
indeed generate all relations.

The best rejected proposal ever

Published March 7, 2007 by lieven

The Oscar in
the category The Best Rejected Research Proposal in Mathematics
(ever) goes to … Alexander Grothendieck
for his proposal Esquisse d’un Programme, Grothendieck\’s research program from 1983, written as
part of his application for a position at the CNRS, the French
equivalent of the NSF. An English translation is
available.

Here is one of the problems discussed :
Give TWO non-trivial elements of
$Gal(\overline{\mathbb{Q}}/\mathbb{Q}) $ the _absolute_
Galois group of the algebraic closure of the rational numbers
$\overline{\mathbb{Q}} $, that is the group of all
$\mathbb{Q} $-automorphisms of $\overline{\mathbb{Q}} $. One element
most of us can give (complex-conjugation) but to find any other
element turns out to be an extremely difficult task.

To get a handle on
this problem, Grothendieck introduced his _’Dessins d’enfants’_
(Children’s drawings). Recall from last session the pictures of the
left and right handed Monsieur Mathieu

The left hand side drawing was associated to a map
$\mathbb{P}^1_{\mathbb{C}} \rightarrow \mathbb{P}^1_{\mathbb{C}} $ which was
defined over the field $\mathbb{Q} \sqrt{-11} $ whereas the right side
drawing was associated to the map given when one applies to all
coefficients the unique non-trivial automorphism in the Galois group
$Gal(\mathbb{Q}\sqrt{-11}/\mathbb{Q}) $ (which is
complex-conjugation). Hence, the Galois group
$Gal(\mathbb{Q}\sqrt{-11}/\mathbb{Q}) $ acts _faithfully_ on the
drawings associated to maps $\mathbb{P}^1_{\mathbb{Q}\sqrt{-11}} \rightarrow
\mathbb{P}^1_{\mathbb{Q}\sqrt{-11}} $ which are ramified only over
the points ${ 0,1,\infty } $.

Grothendieck’s idea was to
extend this to more general maps. Assume that a projective smooth curve
(a Riemann surface) X is defined over the algebraic numbers
$\overline{\mathbb{Q}} $ and assume that there is a map $X
\rightarrow \mathbb{P}^1_{\mathbb{C}} $ ramified only over the points
${ 0,1,\infty } $, then we can repeat the procedure of last time and
draw a picture on X consisting of d edges (where d is the degree
of the map, that is the number of points lying over another point of
$\mathbb{P}^1_{\mathbb{C}} $) between white resp. black points (the
points of X lying over 1 (resp. over 0)).

Call such a drawing a
‘dessin d\’enfant’ and look at the collection of ALL dessins
d’enfants associated to ALL such maps where X runs over ALL curves
defined over $\overline{\mathbb{Q}} $. On this set, there is an action
of the absolute Galois group
$Gal(\overline{\mathbb{Q}}/\mathbb{Q}) $ and if this action would be
faithful, then this would give us insight into this
group. However, at that time even the existence of a map $X \rightarrow
\mathbb{P}^1 $ ramified in the three points ${ 0,1,\infty } $
seemed troublesome to prove, as Grothendieck recalls in his proposal

In more erudite terms, could it be true that
every projective non-singular algebraic curve defined over a number
field occurs as a possible ‚ modular curve‚ parametrising
elliptic curves equipped with a suitable rigidification? Such a
supposition seemed so crazy that I was almost embarrassed to submit
it to the competent people in the domain. Deligne when I consulted
him found it crazy indeed, but didn’t have any counterexample up
his sleeve. Less than a year later, at the International Congress in
Helsinki, the Soviet mathematician Bielyi announced exactly that result,
with a proof of disconcerting simplicity which fit into two little
pages of a letter of Deligne ‚ never, without a doubt, was such a
deep and disconcerting result proved in so few lines!
In
the form in which Bielyi states it, his result essentially says that
every algebraic curve defined over a number field can be obtained as
a covering of the projective line ramified only over the points 0,
1 and infinity. This result seems to have remained more or less
unobserved. Yet, it appears to me to have considerable importance. To
me, its essential message is that there is a profound identity
between the combinatorics of finite maps on the one hand, and the
geometry of algebraic curves defined over number fields on the
other. This deep result, together with the algebraic- geometric
interpretation of maps, opens the door onto a new, unexplored world within reach of all, who pass by without seeing it.

Belyi’s proof is indeed relatively easy
(full details can be found in the paper Dessins d’enfants on the
Riemann sphere by Leila
Schneps). Roughly it goes as follows : as both X and the map are
defined over $\overline{\mathbb{Q}} $ the map is only ramified over
(finitely many) $\overline{\mathbb{Q}} $-points. Let S be the finite
set of all Galois-conjugates of these points and consider the polynomial

$f_0(z_0) = \prod_{s \in S} (z_0 -s) \in
\mathbb{Q}[z_0] $

Now, do a
resultant trick. Consider the
polynomial $f_1(z_1) = Res_{z_0}(\frac{d f_0}{d
z_0},f_0(z_0)-z_1) $ then the roots of $f_1(z_1) $ are exactly the
finite critical values of $f_0 $, $f_1 $ is again defined over
$\mathbb{Q} $ and has lower degree (in $z_1 $) than $f_0 $ in $z_1 $.
Continue this trick a finite number of times untill you have constructed
a polynomial $f_n(z_n) \in \mathbb{Q}[z_n] $ of degree zero.

Composing
the original map with the maps $f_j $ in succession yields that all
ramified points of this composition are
$\mathbb{Q} $-points! Now, we only have to limit the number of
these ramified $\mathbb{Q} $-points (let us call this set T) to three.

Take any three elements of T, then there always exist integers $m,n
\in \mathbb{Z} $ such that the three points go under a linear
fractional transformation (a Moebius-function associated to a matrix in
$PGL_2(\mathbb{Q}) $) to ${ 0,\frac{m}{m+n},1 } $. Under the
transformation $z \rightarrow \frac{(m+n)^{m+n}}{m^m
n^n}z^m(1-z)^n $ the points 0 and 1 go to 0 and
$\frac{m}{m+n} $ goes to 1 whence the ramified points of the
composition are one less in number than T. Continuing in this way we
can get the set of ramified $\mathbb{Q} $-points of a composition at
most having three elements and then a final Moebius transformation gets
them to ${ 0,1,\infty } $, done!

As a tribute for this clever
argument, maps $X \rightarrow \mathbb{P}^1 $ ramified only in 0,1 and
$\infty $ are now called Belyi morphisms. Here is an example of
a Belyi-morphism (and the corresponding dessin d’enfants) associated to
one of the most famous higher genus curves around : the Klein
quartic (if you haven’t done
so yet, take your time to go through this marvelous pre-blog post by
John Baez).

One can define the Klein quartic as the plane projective
curve K with defining equation in
$\mathbb{P}^2_{\\mathbb{C}} $ given by $X^3Y+Y^3Z+Z^3X = 0 $ K has
a large group of automorphism, namely the simple group of order
168 $G = PSL_2(\mathbb{F}_7) =
SL_3(\mathbb{F}_2) $ It is a classical fact (see for example
the excellent paper by Noam Elkies The Klein quartic in number theory) that the quotient map $K \rightarrow K/G =
\mathbb{P}^1_{\mathbb{C}} $ is ramified only in the points
0,1728 and $\infty $ and the number of points of K lying over them
are resp. 56, 84 and 24. Now, compose this map with the Moebius
transormation taking ${ 0,1728,\infty } \rightarrow { 0,1,\infty } $
then the resulting map is a Belyi-map for the Klein quartic. A
topological construction of the Klein quartic is fitting 24 heptagons
together so that three meet in each vertex, see below for the gluing
data-picture in the hyperbolic plane : the different heptagons are given
a number but they appear several times telling how they must fit
together)

The resulting figure has exactly $\frac{7 \times 24}{2} =
84 $ edges and the 84 points of K lying over 1 (the white points in
the dessin) correspond to the midpoints of the edges. There are exactly
$\frac{7 \times 24}{3}=56 $ vertices corresponding to the 56 points
lying over 0 (the black points in the dessin). Hence, the dessin
d\’enfant associated to the Klein quartic is the figure traced out by
the edges on K. Giving each of the 168 half-edges a
different number one assigns to the white points a permutation of order
two and to the three-valent black-points a permutation of order three,
whence to the Belyi map of the Klein quartic corresponds a
168-dimensional permutation representation of $SL_2(\mathbb{Z}) $,
which is not so surprising as the group of automorphisms is
$PSL_2(\mathbb{F}_7) $ and the permutation representation is just the
regular representation of this group.

Next time we will see how
one can always associate to a curve defined over
$\overline{\mathbb{Q}} $ a permutation representation (via the Belyi
map and its dessin) of one of the congruence subgroups $\Gamma(2) $ or
$\Gamma_0(2) $ or of $SL_2(\mathbb{Z}) $ itself.